小编Mis*_*wan的帖子

我是iOS新手我正在使用uitableview单元格我已将按钮拖放到对象检查器的表格视图单元格中.但是当我制作它的IBoutlet时,它显示的错误如"Outlets无法连接到重复内容"这意味着什么？我们不能在tableview cell中创建UI按钮的出口.请仔细检查它.我被困在里面.我正在制作这样一个出口:

@property (weak, nonatomic) IBOutlet UIButton *sa;

并且错误是"UI按钮的出口无效"

我有这个字符串,["658681","655917","655904"]我希望在这个表单中更改此字符串658681,655917,655904如何更改？下面是我的字符串代码

- (IBAction)Searchbtn:(id)sender {

    NSData *data=[NSJSONSerialization dataWithJSONObject:getmessageIDArray options:kNilOptions error:nil];
    _finalIDStr=[[NSString alloc] initWithData:data encoding:NSUTF8StringEncoding];
    NSLog(@"the final ID Str==%@",_finalIDStr);



}

这是我试图在IOS中构建的第一个应用程序,我遇到了一些问题.虽然我在这里读过类似的线程但是我找不到答案.我想在我点击按钮时显示popoverview控制器.但是无法做到.我正在获得错误提及以下问题标题下面是我的文件

.h文件

@property (nonatomic,strong) UIPopoverController *popOver;
@property (nonatomic,strong) SecondViewController *popOverView;

.m文件

- (IBAction)Getcompany:(id)sender {
    SecondViewController *popoverview=[self.storyboard instantiateViewControllerWithIdentifier:@"popover"];
    self.popOver =[[UIPopoverController alloc] initWithContentViewController:popoverview];
    [self.popOver presentPopoverFromRect:sender.frame  inView:self.view permittedArrowDirections:UIPopoverArrowDirectionDown animated:YES];// m getting an error in this line
}

Thanx提前.

我已经在按钮单击操作上编写了以下代码,以便创建ui选择器视图

- (IBAction)selectbtn:(id)sender {
    pickerview = [[UIPickerView alloc] initWithFrame:CGRectMake(5, 40, 300, 300)];
    pickerview.showsSelectionIndicator = YES;
    pickerview.hidden = NO;
    pickerview.delegate = self;
    [self.view addSubview:pickerview];
}

这是我的选择代表

-(NSInteger)numberOfComponentsInPickerView:(UIPickerView *)pickerView
{
    return 1;


}
-(NSInteger)pickerView:(UIPickerView *)pickerView numberOfRowsInComponent:(NSInteger)component
{
    return self.jsonresultarr.count;


}
-(NSString *)pickerView:(UIPickerView *)pickerView titleForRow:(NSInteger)row forComponent:(NSInteger)component
{

//    return [[self.jsonresultarr objectAtIndex:row] objectForKey:@"Company_Id"];
  //  NSObject *companyId = [[self.jsonresultarr objectAtIndex:row] objectForKey:@"Company_Id"];
    NSObject *companyName = [[self.jsonresultarr objectAtIndex:row] objectForKey:@"Company_Name"];
    return [NSString stringWithFormat:@"%@", companyName, nil];




}
-(void)pickerView:(UIPickerView *)pickerView didSelectRow:(NSInteger)row inComponent:(NSInteger)component
{
    self.Textbox.text=[[self.jsonresultarr objectAtIndex:row] objectForKey:@"Company_Name"];
    [self.pickerview removeFromSuperview];


}

我的问题是我想在uipicker视图上显示完成按钮,其功能是在我的文本字段中设置选定的值.我怎么能做到.Thanx提前